41! = 33452526613163807108170062053440751665152000000000

which is 50 digits long, the last 9 of them are 0s. Thus the trailing zeros make up 18% of the entire number.

Find n where n! has the largest possible proportion of trailing 0s.

Prove it.

Observe that 10!=3628800;11!=39916800;12!=479001600

It seems that as we move ahead finding the factorial of maor numbers

the proportion of trailing 0s become more and more less.

Thus the answer we are looking for is less than 10.

So finding out the factorial of individual numbers below 10 we find the answer to be 5! and 6!.